Properties of Matter

Young's Modulus is given by :

$ \mathrm{Y}=\frac{\text { Stress }}{\text { Strain }}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \mathrm{~L} / \mathrm{L}}=\frac{\mathrm{FL}}{\mathrm{~A} \Delta \mathrm{~L}} $

Where:

• F is the applied force (load)

• L is the original length.

• A is the area of cross-section

• $\quad \Delta \mathrm{L}$ is the change in length (stretching)

According to question, both wires are stretched by the same magnitude ( $\Delta \mathrm{L}_{\mathrm{A}}=\Delta \mathrm{L}_{\mathrm{B}}$ ) under the same given load $\left(\mathrm{F}_{\mathrm{A}}=\mathrm{F}_{\mathrm{B}}\right)$.

Therefore, for both wires:

$ Y \propto \frac{L}{A} $

Let $Y_A$ and $Y_B$ be the Young's moduli for wires $A$ and $B$ :

$ \frac{Y_A}{Y_B}=\frac{\left(\frac{L_A}{A_A}\right)}{\left(\frac{L_B}{A_B}\right)}=\frac{L_A}{L_B} \times \frac{A_B}{A_A} $

The given values :

• $\mathrm{L}_{\mathrm{A}}=6.0 \mathrm{~cm}, \mathrm{~L}_{\mathrm{B}}=5.4 \mathrm{~cm}$

• $\mathrm{A}_{\mathrm{A}}=3.0 \times 10^{-5} \mathrm{~m}^2, \mathrm{~A}_{\mathrm{B}}=4.5 \times 10^{-5} \mathrm{~m}^2$

$ \begin{gathered} \frac{Y_A}{Y_B}=\frac{6.0}{5.4} \times \frac{4.5 \times 10^{-5}}{3.0 \times 10^{-5}} \\ \frac{Y_A}{Y_B}=\frac{5}{3}=\frac{x}{3} \Rightarrow x=5 \end{gathered} $

So, the value of x is 5 . Hence, the correct option is (D).

When an object floats in a liquid, the weight of the object is exactly balanced by the buoyant force (up thrust) exerted by the liquid.

Weight of block is $=\mathrm{W}=\mathrm{m} \cdot \mathrm{g}=\left(\rho_{\mathrm{b}} \times \mathrm{V}_{\text {total }}\right) \times \mathrm{g}$

Buoyant force ( $\mathrm{F}_{\mathrm{B}}$ ) acting on a submerged body is equal to the weight of displaced liquid,

$ F_B=\left(\rho_1 V_{\text {submerged }}\right) \times g $

For flotation :

$ \mathrm{W}=\mathrm{F}_{\mathrm{B}} $

$ \rho_{\mathrm{b}} \mathrm{~V}_{\text {total }} \mathrm{g}=\rho_{\mathrm{l}} \mathrm{~V}_{\text {submerged }} \mathrm{g} $

For a cubical block with a uniform cross-sectional area (A):

Total Volume $=\mathrm{V}_{\text {total }}=\mathrm{AH}$

Submerged Volume $\mathrm{V}_{\text {submerged }}=\mathrm{Ah}$ (where h is the submerged height)

$\Rightarrow $ $ \rho_{\mathrm{b}}(\mathrm{AH})=\rho_{\mathrm{l}}(\mathrm{Ah}) \Rightarrow \mathrm{h}=\left(\frac{\rho_{\mathrm{b}}}{\rho_{\mathrm{l}}}\right) \mathrm{H} $

Putting the given values,

$ \rho_{\mathrm{b}}=600 \mathrm{~kg} / \mathrm{m}^3, \rho_{\mathrm{l}}=900 \mathrm{~kg} / \mathrm{m}^3 \text { and } \mathrm{H}=8.0 \mathrm{~cm} $

$\Rightarrow $ $ \mathrm{~h}=\left(\frac{600}{900}\right) \times 8.0 \mathrm{~cm} \approx 5.33 \mathrm{~cm} $

Therefore, the height of the submerged part of the block is approximately 5.3 cm.

Hence, the correct option is (D).

A metal rod of original length $L$ that is free to move. If we heat this rod by a temperature difference $\Delta T$, it will naturally expand.

The change in length ( $\Delta \mathrm{L}$ ) is given by the law of linear thermal expansion :

$ \Delta \mathrm{L}=\mathrm{L} \cdot \alpha \cdot \Delta \mathrm{~T} $

Where $\alpha=$ Coefficient of Linear Expansion

When a wire fixed at both ends is cooled, it contracts. Since the supports are rigid, they prevent this contraction, creating a tension in the wire.

$ \text { Strain } \varepsilon=\frac{\Delta \mathrm{L}}{\mathrm{~L}}=\alpha \Delta \mathrm{T} $

According to Hooke's Law, Stress is proportional to Strain within the elastic limit. The constant of proportionality is Young's Modulus (Y).

$ Y=\frac{\text { Stress }}{\text { Strain }} $

$\Rightarrow $ Stress $(\sigma)=\mathrm{Y} \times$ Strain

$ \sigma=\mathrm{Y} \cdot(\alpha \Delta \mathrm{~T}) $

Stress is defined as Force (F) per unit Area (A).

$ \text { Stress }=\frac{\mathrm{F}}{\mathrm{~A}}=\mathrm{Y} \cdot(\alpha \Delta \mathrm{~T}) $

$\Rightarrow $ $\mathrm{F}=(\mathrm{Y} \alpha \Delta \mathrm{T}) \times \mathrm{A}$

$ \Rightarrow \mathrm{F} \propto \Delta \mathrm{~T} \Rightarrow \frac{\mathrm{~F}_1}{\mathrm{~F}_2}=\frac{\Delta \mathrm{T}_1}{\Delta \mathrm{~T}_2} $

Initial Temperature $=\mathrm{t}_0=27^{\circ} \mathrm{C}$

Final Temperature $=\mathrm{t}_1=-43^{\circ} \mathrm{C}$

$ \Delta \mathrm{T}_1=27-(-43)=70^{\circ} \mathrm{C} $

Initial tension in the wire is, $\mathrm{F}_1=\mathrm{T}$

Final tension in the wire is $\mathrm{F}_2=1.4 \mathrm{~T}$

Let the new final temperature be $t_2$.

$ \Delta \mathrm{T}_2=27-\mathrm{t}_2 $

$\Rightarrow $ $\frac{\mathrm{F}_2}{\mathrm{~F}_1}=\frac{1.4 \mathrm{~T}}{\mathrm{~T}}=\frac{27-\mathrm{t}_2}{70}$

$\Rightarrow $ $1.4=\frac{27-t_2}{70}$

$\Rightarrow $ $27-t_2=1.4 \times 70$

$\Rightarrow $ $27-t_2=98$

$\Rightarrow $$ t_2=27-98 $

$\Rightarrow $$ t_2=-71^{\circ} \mathrm{C} $

Using Stokes' Law, the formula for terminal velocity $\left(v_t\right)$ of a spherical body falling through a viscous fluid is :

$ v_t=\frac{2 r^2(\rho-\sigma) g}{9 \eta} $

Where :

• $r=$ Radius of the sphere $r=\frac{\text { Diameter }}{2}=\frac{2 \mathrm{~mm}}{2}=1 \mathrm{~mm}=0.1 \mathrm{~cm}$.

• $\rho=$ Density of the sphere $=\rho=10.5 \mathrm{~g} / \mathrm{cm}^3$

• $\sigma=$ Density of the fluid, Density of glycerine $=\sigma=1.5 \mathrm{~g} / \mathrm{cm}^3$

• $\mathrm{g}=$ Acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^2=1000 \mathrm{~cm} / \mathrm{s}^2$

• $\eta=$ Coefficient of viscosity $=10$ Poise

1 Poise $=1 \frac{\mathrm{~g}}{\mathrm{~cm} \cdot \mathrm{~s}^3}$, which is the CGS unit of viscosity.

Substituting the values into the terminal velocity equation :

$ v_t=\frac{2 \times(0.1)^2 \times(10.5-1.5) \times 1000}{9 \times 10} $

$\Rightarrow $ $\mathrm{v}_{\mathrm{t}}=\frac{2 \times 0.01 \times 9 \times 1000}{9 \times 10}$

$\Rightarrow $ $ \mathrm{v}_{\mathrm{t}}=2 \mathrm{~cm} / \mathrm{s} $

The sphere accelerates until the viscous drag reaches a point where the upward forces equal the weight of the sphere. At this point, it continues to fall at a constant terminal velocity of $2.0 \mathrm{~cm} / \mathrm{s}$.

Young's Modulus (Y) is defined as the ratio of Stress $(\sigma)$ to Strain $(\varepsilon)$ within the elastic limit :

$ Y=\frac{\text { Stress }}{\text { Strain }} $

For the given graph,

• 1. Y -axis: $\operatorname{Strain}(\varepsilon)$

• 2. X -axis: Stress $(\sigma)$

Since Strain is on the Y -axis and Stress is on the X -axis, the slope of the line is given by :

$ \text { Slope }=\frac{\Delta \mathrm{Y}}{\Delta \mathrm{X}}=\frac{\text { Strain }}{\text { Stress }} $

$ \text { Slope }=\frac{1}{\mathrm{Y}} $

It shows that the slope is inversely proportional to Young's Modulus, i.e., higher slope corresponds to smaller Young's Modulus (material is more elastic) whereas lower slope shows larger Young's Modulus (material is less elastic).

So, the material with the largest Young's modulus will be the line with the smallest slope.

• 1. Material A has the highest slope → smallest Y.

• 2. Material C has the lowest slope → largest Y.

Therefore, material C has the largest Young's modulus. Hence, the correct option is (D).

The height (h) to which a liquid rises in a capillary tube is given by:

$ \mathrm{h}=\frac{2 \mathrm{~S} \cos \theta}{\mathrm{r} \rho \mathrm{~g}} $

Where :

• S is the surface tension of the liquid.

• $\theta$ is the angle of contact.

• $\mathrm{r}$ is the inner radius of the capillary tube.

• $\rho$ is the density of the liquid.

• $g$ is the acceleration due to gravity.

Assuming the angle of contact ( θ) and gravity ( $g$ ) remain constant, the height $h$ depends on variables $S, r$, and $\rho$ as,

$ h \propto \frac{S}{r \rho} $

Taking the natural logarithm of both sides :

$ \ln (h)=\ln (S)-\ln (r)-\ln (\rho)+\text { constant } $

Differentiating this equation,

$ \frac{\Delta \mathrm{h}}{\mathrm{~h}}=\frac{\Delta \mathrm{S}}{\mathrm{~S}}-\frac{\Delta \mathrm{r}}{\mathrm{r}}-\frac{\Delta \rho}{\rho} $

So, the percentage change in height is,

$ \% \Delta \mathrm{~h}=\% \Delta \mathrm{~S}-\% \Delta \mathrm{r}-\% \Delta \rho $

It is given that all three quantities (radius, density, and surface tension) decrease by $1 \%$. Therefore, their percentage changes are negative :

• Change in surface tension, $\% \Delta \mathrm{~S}=-1 \%$

• Change in inner radius, $\% \Delta \mathrm{r}=-1 \%$

• Change in density, $\% \Delta \rho=-1 \%$

Substituting these values into our percentage change equation :

$ \% \Delta \mathrm{~h}=(-1 \%)-(-1 \%)-(-1 \%) $

$\Rightarrow $ $ \% \Delta h=-1 \%+1 \%+1 \%=1 \% $

The positive sign indicates that the height of the liquid in the tube will increase by $1 \%$. Hence, the correct option is (B).

Statement I

Consider an imaginary cube of water out of still water in a container,

This tiny cube has mass, so gravity is pulling it down. If nothing were pushing back up against it, it would fall to the bottom. However, the fluid is still (in static equilibrium). This means the net force on our imaginary cube must be zero.

The water below our imaginary cube must be pushing up on it to counteract gravity. Similarly, the water on the sides is pushing inward to keep it from spreading.

This pushing force per unit area is exactly what pressure is. Therefore, the surrounding fluid exerts pressure on our imaginary surface inside the bulk of the liquid. Pressure is a scalar property that exists at every single point within a fluid, not just where it touches a solid container.

Therefore, the statement I is false.

Statement II

A molecule deep in the interior is completely surrounded by other identical molecules on all sides. The attractive forces pull it equally in every direction. The net force on an interior molecule is zero.

A molecule on the surface has molecules next to it and below it, but practically none above it. Because it is missing half of its neighbours, there is a net inward force pulling the surface molecule down into the bulk of the liquid.

Now, if it is to bring an interior molecule up to the surface to create more surface area. It has to be pulled against that net inward force. And whenever an object moves against an opposing force by external force, then work must be done.

By the Work-Energy Theorem, the work is stored as Potential Energy. Therefore, molecules on the surface possess higher potential energy than those in the bulk interior.

Any system always prefers the lowest possible energy state. Because having surface molecules costs extra energy, a liquid will naturally try to minimize the number of molecules on its surface. It does this by minimizing its overall surface area. This macroscopic tendency of a liquid surface to shrink and act like a stretched elastic membrane is exactly what we call surface tension.

Therefore, the statement II is true.

So, statement I is false, and Statement II is true.

Hence, the correct option is (A).

When a capillary tube is dipped in a liquid, the liquid rises due to surface tension until the upward force of surface tension is balanced by the weight of the liquid column.

The surface tension T acts along the circumference of the meniscus. If $\theta$ is the contact angle, the vertical component of the force is :

$ \mathrm{F}_{\mathrm{up}}=\mathrm{T} \cos \theta \times(2 \pi \mathrm{r}) $

The weight of the liquid column of height h is :

$ W=\left(\pi r^2 h\right) \rho g $

At equilibrium :

$ \mathrm{T} \cos \theta(2 \pi \mathrm{r})=\pi \mathrm{r}^2 \mathrm{~h} \rho \mathrm{~g} $

$\Rightarrow $ $ \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{~g}} $

If we assume the two liquids refers to the same substance, then:

• $\mathrm{T}_1=\mathrm{T}_2 \ldots$ (i) (Surface tension is a property of the material and temperature).

• $\rho$ and $g$ are constant.

• $\theta$ is constant.

This results the relationship: $\mathrm{h} \propto \frac{1}{\mathrm{r}}$.

We are given that $r_1>r_2 \Rightarrow \frac{r_1}{r_2}>1$.

So, the ratio of heights of capillary rise,

$ \frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{\mathrm{r}_2}{\mathrm{r}_2}<1 \Rightarrow \mathrm{~h}_1<\mathrm{h}_2 \ldots (ii)$

From both the relations, $\mathrm{h}_1<\mathrm{h}_2$ and $\mathrm{T}_1=\mathrm{T}_2$.

Hence, the correct option is (B).

To find the final length of the composite rod, we need to calculate the expansion of both the aluminium and steel sections individually and then add them to the original length.

Total Initial Length $\left(\mathrm{L}_1\right)$ is 120 cm at $30^{\circ} \mathrm{C}$.

Since both rods have the same length, the initial length of each rod is, $\mathrm{L}_{\mathrm{Al}}=\frac{120}{2}=60 \mathrm{~cm}$ and $\mathrm{L}_{\mathrm{St}}=\frac{120}{2}=$ 60 cm .

The change in temperature is,

$ \Delta \mathrm{T}=\mathrm{T}_{\text {final }}-\mathrm{T}_{\text {initial }}=100^{\circ} \mathrm{C}-30^{\circ} \mathrm{C}=70^{\circ} \mathrm{C} $

Coefficients of linear expansion ( $\alpha$ ) for aluminium is $\alpha_{\mathrm{Al}}=24 \times 10^{-6} /{ }^{\circ} \mathrm{C}$ and that for steel is $\alpha_{\mathrm{St}}=1.2 \times 10^{-5} /{ }^{\circ} \mathrm{C}=12 \times 10^{-6} /{ }^{\circ} \mathrm{C}$.

The linear expansion is $\Delta \mathrm{L}=\mathrm{L}_{\text {initial }} \cdot \alpha \cdot \Delta \mathrm{T}$

For Aluminium :

$ \Delta \mathrm{L}_{\mathrm{Al}}=60 \times\left(24 \times 10^{-6}\right) \times 70 $

$\Rightarrow $ $\Delta \mathrm{L}_{\mathrm{Al}}=60 \times 70 \times 24 \times 10^{-6}$

$\Rightarrow $ $ \Delta \mathrm{L}_{\mathrm{Al}}=4200 \times 24 \times 10^{-6}=100800 \times 10^{-6}=0.1008 \mathrm{~cm} $

For Steel :

$ \Delta \mathrm{L}_{\mathrm{St}}=60 \times\left(12 \times 10^{-6}\right) \times 70 $

$\Rightarrow $ $ \Delta \mathrm{L}_{\mathrm{St}}=4200 \times 12 \times 10^{-6}=50400 \times 10^{-6}=0.0504 \mathrm{~cm} $

The total change in length ( $\Delta \mathrm{L}_{\text {total }}$ ) is :

$ \Delta \mathrm{L}_{\text {total }}=\Delta \mathrm{L}_{\mathrm{Al}}+\Delta \mathrm{L}_{\mathrm{St}} $

$\Rightarrow $ $ \Delta \mathrm{L}_{\text {total }}=0.1008+0.0504=0.1512 \mathrm{~cm} $

So, the final length of the rod is :

$ \mathrm{L}_{\mathrm{final}}=120+0.1512=120.1512 \mathrm{~cm} $

$\Rightarrow $ $ \mathrm{L}_{\mathrm{final}} \approx 120.15 \mathrm{~cm} $

Therefore, the final length of the combination is 120.15 cm .

Hence, the correct option is (C).

Applying the equation of continuity

The rate of flow ( Q ) is constant throughout the tube.

$ \mathrm{Q}=\mathrm{A}_1 \mathrm{v}_1=\mathrm{A}_2 \mathrm{v}_2 $

Where :

• $\mathrm{A}_1=$ cross-section area at $\mathrm{A}, \mathrm{A}_1=6 \mathrm{~cm}^2$

• $\mathrm{A}_2=$ cross-section area at $\mathrm{B}, \mathrm{A}_2=3 \mathrm{~cm}^2$

• $\mathrm{V}_1, \mathrm{~V}_2$ are velocities at A and B .

$ 6 v_1=3 v_2 \Rightarrow v_2=2 v_1 $

Applying Bernoulli's principle between two points,

$ P_1+\frac{1}{2} \rho v_1^2+\rho g h_1=P_2+\frac{1}{2} \rho v_2^2 $

For a horizontal tube, the height is the same, $\mathrm{h}_1=\mathrm{h}_2$

$ P_1-P_2=\frac{1}{2} \rho\left(v_2^2-v_1^2\right) $

The pressure difference ( $P_1-P_2$ ) is measured by the difference in the heights of the water columns ( $h$ ) :

$ \mathrm{P}_1-\mathrm{P}_2=\rho \mathrm{gh} $

From relations (ii) and (iii),

$ \rho g h=\frac{1}{2} \rho\left(v_2^2-v_1^2\right) $

$\Rightarrow $ $ 2 g h=v_2^2-v_1^2 $

Substituting $\mathrm{v}_2=2 \mathrm{v}_1$ from relation (i) :

$ 2 g h=\left(2 v_1\right)^2-v_1^2 $

$\Rightarrow $ $2 g h=4 v_1^2-v_1^2=3 v_1^2$

$ \Rightarrow v_1=\sqrt{2 g h / 3} $

The given values are,

• $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2=1000 \mathrm{~cm} / \mathrm{s}^2$

• $\mathrm{h}=5 \mathrm{~cm}$

Putting the values,

$ v_1=\sqrt{\frac{2 \times 1000 \times 5}{3}}=\sqrt{\frac{10000}{3}}=\frac{100}{\sqrt{3}} \mathrm{~cm} / \mathrm{s} $

So, the rate of flow (Q),

$ Q=A_1 \cdot v_1 $

$\Rightarrow $ $ Q=6 \cdot \frac{100}{\sqrt{3}}=2 \times \sqrt{3} \times \sqrt{3} \times \frac{100}{\sqrt{3}}=200 \sqrt{3} \mathrm{~cm}^3 / \mathrm{s} $

Therefore, the rate of flow of water through the horizontal tube is $200 \sqrt{3} \mathrm{~cm}^3 / \mathrm{s}$.

Hence, the correct option is (D).

Let’s break down the solution step by step:

First, note the given values:

Length of the wire, $L = 3 \, \text{m}$

Radius of the wire, $r = 3 \, \text{mm} = 3 \times 10^{-3} \, \text{m}$

Extension of the wire, $\Delta L = 0.1 \, \text{mm} = 1 \times 10^{-4} \, \text{m}$

Mass attached, $m = 50 \, \text{kg}$

Acceleration due to gravity, $g = 3\pi \, \text{m/s}^2$

The force (weight) acting on the wire is:

$F = mg = 50 \times 3\pi = 150\pi \, \text{N}$

Next, calculate the cross-sectional area, $A,$ of the wire:

$A = \pi r^2 = \pi (3 \times 10^{-3})^2 = \pi \times 9 \times 10^{-6} = 9\pi \times 10^{-6} \, \text{m}^2$

Young’s modulus, $E,$ is given by:

$E = \frac{FL}{A \Delta L}$

Substitute the values into the formula:

$ E = \frac{(150\pi) \times 3}{(9\pi \times 10^{-6}) \times (1 \times 10^{-4})} $

Simplify step by step:

Multiply in the numerator:

$150\pi \times 3 = 450\pi$

Multiply in the denominator:

$9\pi \times 10^{-6} \times 1 \times 10^{-4} = 9\pi \times 10^{-10}$

Cancel the common factor $\pi$ in the numerator and denominator:

$ E = \frac{450}{9 \times 10^{-10}} $

Divide 450 by 9:

$ E = \frac{450}{9} \times \frac{1}{10^{-10}} = 50 \times 10^{10} = 5 \times 10^{11} \, \text{Nm}^{-2} $

The problem states that Young's modulus can be expressed as $P \times 10^{11} \, \text{Nm}^{-2}.$ Here we have:

$ P = 5 $

Thus, the correct option is:

Option D (5).

$\begin{aligned} & \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}}=\frac{2 \times 70 \times 1}{1 \times 980 \times 10^{-2}} \\ & \mathrm{~h}=\frac{100}{7} \mathrm{~cm} \\ & \sin 60^{\circ}=\frac{\mathrm{h}}{\ell} \\ & \ell=\frac{\mathrm{h} \times 2}{\sqrt{3}} \\ & \ell=\frac{100}{7} \times \frac{2}{\sqrt{3}} \\ & =\frac{200}{7 \times \sqrt{3}} \\ & =16.49 \mathrm{~cm} \end{aligned}$

Given:

The ratio of lengths: $\frac{L_A}{L_B} = \frac{1}{3}$

The ratio of diameters: $\frac{d_A}{d_B} = 2$

Both wires are subject to the same stretching force, and since they are made of the same material, their Young's modulus ($Y$) is the same.

The elongation ($\Delta L$) of a wire subject to a force is given by:

$ \Delta L = \frac{F \cdot L}{A \cdot Y} $

where $F$ is the force applied, $L$ is the original length, $A$ is the cross-sectional area, and $Y$ is Young's modulus.

For wires $A$ and $B$:

$\Delta L_A = \frac{F_A \cdot L_A}{A_A \cdot Y_A}$

$\Delta L_B = \frac{F_B \cdot L_B}{A_B \cdot Y_B}$

Since $F_A = F_B$ and $Y_A = Y_B$, the ratio of their elongations becomes:

$ \frac{\Delta L_A}{\Delta L_B} = \frac{L_A \cdot A_B}{L_B \cdot A_A} $

The cross-sectional area $A$ is a function of diameter, $A = \frac{\pi}{4} d^2$. Therefore,

$ A_A = \frac{\pi}{4} d_A^2 \quad \text{and} \quad A_B = \frac{\pi}{4} d_B^2 $

Substituting these into the elongation ratio:

$ \frac{\Delta L_A}{\Delta L_B} = \left(\frac{L_A}{L_B}\right) \left(\frac{\frac{\pi}{4} d_B^2}{\frac{\pi}{4} d_A^2}\right) = \left(\frac{L_A}{L_B}\right) \left(\frac{d_B}{d_A}\right)^2 $

Substitute the given ratios:

$ \frac{\Delta L_A}{\Delta L_B} = \left(\frac{1}{3}\right)\left(\frac{1}{2}\right)^2 = \left(\frac{1}{3}\right)\left(\frac{1}{4}\right) = \frac{1}{12} $

Thus, the ratio of the elongations of wires $A$ and $B$ is $\frac{1}{12}$.

$\begin{aligned} & \text { Shear moduli }=\frac{\sigma_{\text {shear }}}{\theta} \\ & 10^{10}=\frac{10^5}{\pi \times 16 \times 10^{-4}} \times \frac{1}{\theta} \\ & \theta=\frac{1}{160 \pi} \text { Radian } \end{aligned}$

The contact angle, $\theta,$ of a liquid in a capillary tube determines its meniscus shape:

If $\theta < 90^\circ,$ then $\cos \theta > 0$ and the liquid wets the tube, giving a concave meniscus.

If $\theta > 90^\circ,$ then $\cos \theta < 0$ and the liquid is non-wetting, resulting in a convex meniscus.

If $\theta = 90^\circ,$ then $\cos \theta = 0$ and the meniscus is essentially flat.

We are given the ratio:

$K = \frac{\cos \theta_A}{\cos \theta_B}.$

For $K$ to be negative:

The numerator and denominator must have opposite signs.

This means one of the liquids has $\cos \theta > 0$ (concave meniscus) and the other has $\cos \theta < 0$ (convex meniscus).

Consider Option D:

It states that if $K$ is negative, then liquid A has a concave meniscus (so $\theta_A < 90^\circ$, $\cos \theta_A > 0$) and liquid B has a convex meniscus (so $\theta_B > 90^\circ$, $\cos \theta_B < 0$).

Hence, the ratio becomes negative:

$K = \frac{(+)}{(-)} < 0.$

Therefore, the correct statement is:

Option D

First, when the two vessels are connected, the water levels equalize at

$H = \frac{H_1 + H_2}{2} = \frac{10 + 6}{2} = 8\text{ m}.$

We can get the work done by gravity from the loss of gravitational potential energy:

Initial potential energy (taking the bottom as zero)

For a column of height $H$ and cross-section $A$,

$U = \rho g A\int_0^H z\,dz = \frac{\rho g A H^2}{2}.$

So

$U_i = \frac{\rho g A}{2}\bigl(H_1^2 + H_2^2\bigr) = \frac{10^3\cdot 10\cdot 2}{2}(10^2 + 6^2) = 10^4\,(100 + 36) = 1.36\times10^6\text{ J}.$

Final potential energy (both at height 8 m)

$U_f = 2\;\frac{\rho g A H^2}{2}\;\Big|_{H=8} = \rho g A\,(8^2) = 10^3\cdot10\cdot2\cdot64 = 1.28\times10^6\text{ J}.$

Work done by gravity = drop in potential energy

$W = U_i - U_f = 1.36\times10^6 - 1.28\times10^6 = 0.08\times10^6 = 8\times10^4\text{ J}.$

Answer: 8×10^4 J (Option C).

To determine the viscosity of the oil, we use the formula for the terminal velocity of a sphere falling through a viscous fluid:

$ v_T = \frac{2}{9} \frac{(\rho_s - \rho_f) r^2 g}{\eta} $

Where:

$ v_T $ is the terminal velocity,

$ \rho_s $ is the density of the steel ball,

$ \rho_f $ is the density of the fluid,

$ r $ is the radius of the ball,

$ g $ is the acceleration due to gravity,

$ \eta $ is the viscosity of the fluid.

Given:

Diameter of the steel ball = 3.6 mm, therefore radius $ r = 1.8 $ mm = $ 1.8 \times 10^{-3} $ m,

Terminal velocity $ v_T = 2.45 \times 10^{-2} $ m/s,

Density of oil $ \rho_f = 925 $ kg/m$^3$,

Density of steel $ \rho_s = 7825 $ kg/m$^3$,

Acceleration due to gravity $ g = 9.8 $ m/s$^2$.

Rearranging the formula to solve for $\eta$:

$ \eta = \frac{2}{9} \cdot \frac{(\rho_s - \rho_f) \cdot r^2 \cdot g}{v_T} $

Substituting the known values into the equation:

$ \eta = \frac{2}{9} \cdot \frac{(7825 - 925) \cdot (1.8 \times 10^{-3})^2 \cdot 9.8}{2.45 \times 10^{-2}} $

Calculating this, we approximate the viscosity $\eta$ to be:

$ \eta \approx 1.99 \, \text{Pa}\cdot\text{s} $

Thus, the viscosity of the oil is approximately 1.99 Pa·s.

Apply Bernouli equation between points 1 & 2

$ \begin{aligned} & P_1+\frac{1}{2} \rho v_1^2+\rho g h=P_2+\frac{1}{2} \rho v_2^2+0 \\ & P_0+\frac{m g}{A}+\rho g \frac{70}{100}=P_0+\frac{1}{2} \rho v_2^2 \\ & \frac{5000}{0.5}+10^3 \times 10 \frac{70}{100}=\frac{1}{2} \times 10^3 v_2^2 \\ & 10^3+10^3 \times 7=\frac{10^3}{2} v_2^2 \\ & v_2^2=16 \\ & v_2=4 \mathrm{~m} / \mathrm{s} \end{aligned}$

As the tank area is large $v_1$ is negligible compared $\text { to } v_2$

To determine the surface energy released when two identical water drops coalesce to form a larger drop, let's go through the process step by step:

Volume Conservation:

When two droplets, each with a radius $ R $, merge to form a larger droplet, the total volume is conserved. Therefore:

$ 2 \times \frac{4}{3} \pi R^3 = \frac{4}{3} \pi r^3 $

Solving for $ r $, the radius of the larger drop, we find:

$ r = 2^{1/3} R $

Initial Surface Energy ($ U_i $):

The initial surface energy of the two smaller droplets is given by the surface area of each multiplied by the surface tension $ T $:

$ U_i = 2 \times 4 \pi R^2 T $

Final Surface Energy ($ U_f $):

The surface energy of the larger droplet is given by:

$ U_f = 4 \pi r^2 T = 4 \pi R^2 T (2^{2/3}) $

Energy Released:

The energy released in the process is the difference between the initial and final surface energy:

$ \text{Heat lost} = U_i - U_f = 4 \pi R^2 T \left[2 - 2^{2/3}\right] $

This expression gives us the amount of surface energy that is released when two identical water drops combine to form a single larger drop.

The fractional compression $\left( \frac{\Delta V}{V} \right)$ of water at a depth of 2.5 km below sea level is calculated as follows:

Given:

Bulk modulus of water, $B = 2 \times 10^9 \, \text{N/m}^2$

Density of water, $\rho = 10^3 \, \text{kg/m}^3$

Acceleration due to gravity, $g = 10 \, \text{m/s}^2$

The relationship between pressure change and volume change using the bulk modulus is given by:

$ B = \frac{\rho gh}{\left(\frac{\Delta V}{V}\right)} $

Rearranging the formula to solve for the fractional compression:

$ \frac{\Delta V}{V} \times 100 = \frac{\rho gh}{B} \times 100 $

Substituting in the given values:

$ \frac{1000 \times 10 \times 2.5 \times 10^3}{2 \times 10^9} \times 100\% $

Calculating the result gives:

$ = 1.25\% $

Thus, the fractional compression of water at this depth is 1.25%.

First, let’s find the weight of the cube and set it equal to the upward buoyant force that keeps the cube floating.

Step 1: Set up the equation for floating
The weight of the cube (mass × gravity) is balanced by the buoyant force (density of water × volume of cube under water × gravity):
$\mathrm{Mg} = \mathrm{F}_{\mathrm{B}} \Rightarrow (400 \times 10^{-3}) = 10^3 \times \mathrm{V}_{\mathrm{d}}$

Step 2: Solve for the volume under water
We find the volume of the cube under water ($\mathrm{V}_\mathrm{d}$):
$\mathrm{V}_{\mathrm{d}} = 400 \times 10^{-6}~\mathrm{m}^3$

Step 3: Find the total volume of the cube
Each edge of the cube is 10 cm, so total volume is:
$ (10 \times 10^{-2})^3 $

Step 4: Find the volume outside water
Subtract the volume under water from the total volume to get the volume outside water:
$(\text {Vol.})_{\text{outside}} = (10 \times 10^{-2})^3 - 400 \times 10^{-6}$

Calculate the answer:
$= 600 \times 10^{-6}~\mathrm{m}^3 = 600~\mathrm{cm}^3$

We know, terminal velocity of a sphere of radius R in a liquid of viscosity $\eta$,

$v = {2 \over 9}{{{R^2}} \over \eta }(\sigma - \rho )$ .... (1)

where, $\sigma$ = mass of density of sphere

$\rho$ = density of liquid

we can see, $v \propto {R^2}$ (for constant $\eta,\sigma$ & $\rho$)

Hence, graph between v and R is parabola.

As v depends on R so the terminal velocities of different diameter balls will be different.

We know, the viscosity of a liquid usually decreases as the temperature increases and $v \propto {1 \over \eta }$

So terminal velocity depends on the temperature. $T \uparrow \Rightarrow \eta \downarrow \Rightarrow v \uparrow $

As the equation $v = {2 \over 9}{{{R^2}} \over v}(\sigma - \rho )$ involves density of liquid $\rho$. So the experiment can be utilized to asses it.

From (1), $\eta = {2 \over 9}{{{R^2}} \over v}(\sigma - \rho )$

Here, $\eta$ does not depend on initial speed of the sphere. Hence, option 3 is correct.

(A) Surface tension arises due to the extra potential energy of the molecules at the surface of a liquid compared to the molecules in the interior.

(B) When a liquid is heated, the kinetic energy of it molecules increases, causing them to move more freely and overcome the intermolecular forces that contribute to viscosity, thus lowering it.

(C) Higher temperature in gas means faster moving molecules, resulting in more collisions and therefore an increase in viscosity.

(D) Reynold's number $ = {{Inertial\,force} \over {Visous\,force}}$

${R_e} = {{\rho VD} \over \mu }$,

v = velocity of flow

D = Diameter of pipe

If, $R_e < 2000$, flow is laminar flow

$2000 < R_e < 3000 \Rightarrow$ transition region

$R_e > 3000 \Rightarrow$ turbulant flow.

(E) Tangent at a point to the streamline gives the direction of net velocity of the flow. If the two streamlines intersect each other. It signifies two directions of velocity which cannot be possible. Thus, two streamlines cannot intersect each other. Hence, option 2 is correct.

$ P_{\text{bubble}} = P_{\text{atm}} + \rho g h + \frac{2S}{r} $

Given that the pressure inside the bubble is $P_{\text{atm}} + 2100\,\text{N/m}^2$ and the liquid pressure at a depth of 20 cm is

$ P_{\text{liquid}} = P_{\text{atm}} + \rho g h = P_{\text{atm}} + 1000 \times 10 \times 0.20 = P_{\text{atm}} + 2000\,\text{N/m}^2, $

the pressure difference due solely to surface tension is

$ \Delta P = \Bigl(P_{\text{bombubble}} - P_{\text{liquid}}\Bigr) = \bigl(P_{\text{atm}}+2100\bigr) - \bigl(P_{\text{atm}}+2000\bigr) = 100\,\text{N/m}^2. $

The Laplace pressure difference for a bubble (which has one interface) is given by

$ \Delta P = \frac{2S}{r}. $

The given bubble radius is 0.1 cm, which in SI units is

$ r = 0.1\,\text{cm} = 0.001\,\text{m}. $

Substituting the known values into the Laplace equation gives

$ \frac{2S}{0.001} = 100. $

Solving for $S$:

$ 2S = 100 \times 0.001 = 0.1, $

$ S = \frac{0.1}{2} = 0.05\,\text{N/m}. $

Thus, the surface tension of the liquid is

$ \boxed{0.05\,\text{N/m}}. $

To solve the problem, we need to compare the increase in surface area when the big drop is split into small drops.

Step 1: Determine the radius of the small drops

For 27 small drops:

Total volume is conserved:

$ 27 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies r^3 = \frac{R^3}{27} \implies r = \frac{R}{3}. $

For 64 small drops:

Similarly,

$ 64 \times \frac{4}{3}\pi r'^3 = \frac{4}{3}\pi R^3 \implies r'^3 = \frac{R^3}{64} \implies r' = \frac{R}{4}. $

Step 2: Calculate the surface area before and after the break-up

Initial surface area of the big drop:

$ A_{\text{initial}} = 4\pi R^2. $

For 27 drops:

Surface area of one small drop:

$ 4\pi\left(\frac{R}{3}\right)^2 = \frac{4\pi R^2}{9}. $

Total surface area:

$ A_{27} = 27 \times \frac{4\pi R^2}{9} = 3 \times 4\pi R^2 = 12\pi R^2. $

Increase in surface area:

$ \Delta A_{27} = 12\pi R^2 - 4\pi R^2 = 8\pi R^2. $

For 64 drops:

Surface area of one small drop:

$ 4\pi\left(\frac{R}{4}\right)^2 = 4\pi \frac{R^2}{16} = \frac{\pi R^2}{4}. $

Total surface area:

$ A_{64} = 64 \times \frac{\pi R^2}{4} = 16\pi R^2. $

Increase in surface area:

$ \Delta A_{64} = 16\pi R^2 - 4\pi R^2 = 12\pi R^2. $

Step 3: Relate work done to the change in surface area

Since the work done is proportional to the increase in surface area:

$ \text{Work} \propto \Delta A. $

We know that breaking into 27 drops requires 10 J corresponding to an increase of $8\pi R^2$.

Thus, if $W$ is the work done,

$ W_{27} = k \cdot 8\pi R^2 = 10\, \text{J}, $

where $k$ is the proportionality constant.

For 64 drops:

$ W_{64} = k \cdot 12\pi R^2. $

Dividing the two equations:

$ \frac{W_{64}}{10} = \frac{12\pi R^2}{8\pi R^2} = \frac{12}{8} = \frac{3}{2}. $

Thus,

$ W_{64} = 10 \times \frac{3}{2} = 15\, \text{J}. $

Final Answer: 15 J (Option D)

We can solve this using Hooke's Law, which states:

$F = kx,$

where:

• $F$ is the force (tension),

• $k$ is the spring constant, and

• $x$ is the extension of the spring.

Follow these steps:

For the first scenario (extension $x_1$ under 5 N):

$5 = kx_1 \quad \Rightarrow \quad k = \frac{5}{x_1}.$

For the second scenario (extension $x_2$ under 7 N):

$7 = kx_2 \quad \Rightarrow \quad k = \frac{7}{x_2}.$

Equate the two expressions for $k$:

$\frac{5}{x_1} = \frac{7}{x_2} \quad \Rightarrow \quad \frac{x_1}{x_2} = \frac{5}{7} \quad \Rightarrow \quad x_2 = \frac{7}{5}x_1.$

To find the tension for the extension of $\left(5x_1 - 2x_2\right)$, use Hooke's Law again:

$\text{Tension, } F = k \left(5x_1 - 2x_2\right).$

Substitute $k = \frac{5}{x_1}$:

$F = \frac{5}{x_1} \left(5x_1 - 2x_2\right).$

Substitute the expression for $x_2$:

$F = \frac{5}{x_1} \left(5x_1 - 2\left(\frac{7}{5}x_1\right)\right) = \frac{5}{x_1} \left(5x_1 - \frac{14}{5}x_1\right).$

Simplify the expression inside the parentheses:

$5x_1 - \frac{14}{5}x_1 = \left(\frac{25}{5} - \frac{14}{5}\right)x_1 = \frac{11}{5}x_1.$

Now substitute back:

$F = \frac{5}{x_1} \cdot \frac{11}{5}x_1 = 11 \text{ N}.$

Thus, the tension in the spring for the extension $\left(5x_1-2x_2\right)$ is $11 \text{ N}.$

The correct option is Option C.

When water flows from a region of higher pressure to a region of lower pressure, the difference in pressure is converted into kinetic energy of the flowing water. According to Bernoulli's principle for horizontal flow (where gravitational effects can be ignored), this conversion can be expressed as:

$ \frac{1}{2}\rho v^2 = P_1 - P_2 $

Here,

$\rho$ is the density of water,

$v$ is the speed of the water,

$P_1$ is the initial pressure when the valve is closed,

$P_2$ is the pressure after the valve is opened.

By solving for $v$, we have:

$ v = \sqrt{\frac{2(P_1 - P_2)}{\rho}} $

This equation shows that the speed of the water, $v$, is proportional to the square root of the pressure difference:

$ v \propto \sqrt{P_1 - P_2} $

Thus, the correct relationship is given by the square root of the pressure difference.

Let's analyze the statements one by one:

$\textbf{Statement I: "The hot water flows faster than cold water."}$

When water is heated, its viscosity decreases. In simple terms, hot water is "thinner" than cold water.

Lower viscosity means there is less internal friction in the fluid, which allows it to flow more easily and, consequently, faster under the same conditions.

Therefore, under identical pressure conditions, hot water indeed flows faster than cold water.

This makes Statement I true.

$\textbf{Statement II: "Soap water has higher surface tension as compared to fresh water."}$

Soap is a surfactant. Surfactants reduce the surface tension of the liquid by interfering with the cohesive forces between the water molecules.

Fresh (pure) water has a relatively high surface tension (about 72.8 mN/m at room temperature), but when soap is added, the surface tension is decreased.

Therefore, soap water has lower, not higher, surface tension compared to pure water.

This makes Statement II false.

Based on this analysis, the correct answer is:

Option D: Statement I is true but Statement II is false.

$\begin{aligned} & \mathrm{A}_1 \mathrm{v}_1=\mathrm{A}_2 \mathrm{~V}_2 \Rightarrow 2 \pi(2 \mathrm{R})^2=\mathrm{V}_2 \pi \mathrm{R}^2 \\ & \therefore \mathrm{~V}_2=8 \mathrm{~m} / \mathrm{s} \end{aligned}$

$\begin{aligned} & \mathrm{mg}-\mathrm{F}_{\mathrm{B}}-\mathrm{f}=0 \\ & \Rightarrow \mathrm{mg}-\frac{\mathrm{mg}}{2}-\mathrm{f}=0 \\ & \therefore \mathrm{f}=\frac{\mathrm{mg}}{2} \end{aligned}$

To find the ratio between the volumes of the first and the second soap bubble, we need to understand the relation between the excess pressure inside a soap bubble and its volume.

The excess pressure ($P$) inside a soap bubble is given by the formula:

$P = \frac{4T}{r}$

where $T$ is the surface tension of the soap solution, and $r$ is the radius of the soap bubble. For a soap bubble, the factor of 4 comes from having two surfaces (inner and outer), each contributing $2T/r$ to the pressure.

Given that the excess pressure inside the first soap bubble ($P_1$) is thrice the excess pressure inside the second soap bubble ($P_2$), we can write:

$P_1 = 3P_2$

Substituting the formula for excess pressure, we get:

$\frac{4T}{r_1} = 3 \times \frac{4T}{r_2} \Rightarrow \frac{1}{r_1} = 3 \times \frac{1}{r_2} \Rightarrow \frac{r_2}{r_1} = 3$

Next, we calculate the ratio between their volumes. The volume ($V$) of a sphere (or a bubble) is given by:

$V = \frac{4}{3}\pi r^3$

Thus, the volume ratio of the first bubble ($V_1$) to the second bubble ($V_2$) is:

$\frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \left(\frac{r_1}{r_2}\right)^3$

Since we found that $\frac{r_2}{r_1} = 3$, it then follows that:

$\frac{V_1}{V_2} = \left(\frac{1}{3}\right)^3 = \frac{1}{27}$

Therefore, the correct option is:

Option B: $1: 27$

To solve this problem, we can use the concepts of terminal velocity and the forces acting on the spherical ball. First, let's analyze the situation step-by-step.

When the ball falls freely under gravity, it achieves a terminal velocity $v_t$ in water. This terminal velocity is reached when the gravitational force is balanced by the drag force and the buoyant force in the water.

The forces acting on the ball are:

1. Gravitational Force: $F_g = mg$

2. Buoyant Force: $F_b = \rho_{\text{water}} V g$

3. Drag Force: $F_d = 6 \pi \eta r v_t$

Where,

$m$ is the mass of the ball.

$g$ is the acceleration due to gravity ($9.8 \, \mathrm{m/s^2}$).

$\rho_{\text{water}}$ is the density of water ($1000 \, \mathrm{kg/m^3}$).

$V$ is the volume of the ball ($\frac{4}{3} \pi r^3$).

$\eta$ is the coefficient of viscosity of water ($9.8 \times 10^{-6} \, \mathrm{Ns/m^2}$).

$r$ is the radius of the ball ($1 \times 10^{-4} \, \mathrm{m}$).

$v_t$ is the terminal velocity.

Using the equilibrium condition at terminal velocity:

$F_g = F_b + F_d$

$mg = \rho_{\text{water}} V g + 6 \pi \eta r v_t$

First, compute the mass of the ball:

$m = \rho_{\text{ball}} \times V = \rho_{\text{ball}} \times \frac{4}{3} \pi r^3$

$m = 10^5 \, \mathrm{kg/m^3} \times \frac{4}{3} \pi (1 \times 10^{-4} \, \mathrm{m})^3$

$m = 10^5 \,\mathrm{kg/m^3} \times \frac{4}{3} \pi \times 10^{-12} \,\mathrm{m^3}$

$m = \frac{4}{3} \pi \times 10^{-7} \, \mathrm{kg}$

Next, solve for the terminal velocity $v_t$ using the equilibrium equation:

$mg = \rho_{\text{water}} \frac{4}{3} \pi r^3 g + 6 \pi \eta r v_t$

$v_t = \frac{mg - \rho_{\text{water}} \frac{4}{3} \pi r^3 g}{6 \pi \eta r}$

$v_t = \frac{\frac{4}{3} \pi \times 10^{-7} \times 9.8 - 1000 \times \frac{4}{3} \pi (1 \times 10^{-4})^3 \times 9.8}{6 \pi \times 9.8 \times 10^{-6} \times 10^{-4}}$

$v_t = \frac{\frac{4}{3} \pi \times 10^{-7} \times 9.8 - 1000 \times \frac{4}{3} \pi \times 10^{-12} \times 9.8}{6 \pi \times 9.8 \times 10^{-10}}$

$v_t = \frac{\frac{4}{3} \pi \times 9.8 \times 10^{-7} (1 - 10^{-5})}{6 \pi \times 9.8 \times 10^{-10}}$

$v_t = \frac{\frac{4}{3} \times 10^{-7}}{6 \times 10^{-10}}$

$v_t = \frac{4}{18} \times 10^3 \, \mathrm{m/s}$

$v_t \approx 222.22 \, \mathrm{m/s}$

The height $h$ required to reach this terminal velocity while the ball falls freely under gravity can be found using the kinematic equation:

$v_t = \sqrt{2gh}$

$h = \frac{v_t^2}{2g}$

$h = \frac{(222.22)^2}{2 \times 9.8}$

$h = \frac{49328.88}{19.6}$

$h \approx 2517.8 \, \mathrm{m}$

So, the closest value of $h$ is approximately:

Option A: 2518 m.

To solve this problem, we consider the buoyancy and weight force acting on the sphere. For the sphere to just float on water, the weight of the water displaced by the sphere must be equal to the weight of the sphere. The volume of water displaced by the sphere is equivalent to the outer volume of the sphere minus the volume of the cavity inside it.

The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$, where $r$ is the radius of the sphere. For the given sphere, its outer radius is $R = \frac{D}{2}$, and the radius of the cavity is $r = \frac{d}{2}$. Therefore, the volume of the sphere excluding the cavity is:

$V_{\text{solid part}} = \frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3$

$V_{\text{solid part}} = \frac{4}{3}\pi \left(\frac{D}{2}\right)^3 - \frac{4}{3}\pi \left(\frac{d}{2}\right)^3$

Relative density ($\sigma$) is defined as the ratio of the density of an object to the density of water. This means the actual density of the sphere is $\sigma \times \text{density of water}$. Since the object just floats, the weight of the displaced water is equal to the weight of the solid part of the sphere (ignoring the cavity), which can be mathematically represented as:

$\text{Weight of solid part} = \text{Weight of displaced water}$

$\sigma \cdot \rho_{\text{water}} \cdot V_{\text{solid part}} \cdot g = \rho_{\text{water}} \cdot V_{\text{displaced water}} \cdot g$

Since the sphere is floating, $V_{\text{displaced water}} = \frac{4}{3}\pi R^3$, the equation simplifies to:

$\sigma \left(\frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3$

Cancelling out common terms gives:

$\sigma (R^3 - r^3) = R^3$

Given that $\sigma$ is the relative density, we can rearrange the equation to solve for the ratio of $D/d$ or equivalently $R/r$:

$\sigma = \frac{R^3}{R^3 - r^3}$

Solving for $R/r$:

$R^3 (1 - \sigma) + \sigma r^3 = 0$

$\frac{R^3}{r^3} = \frac{\sigma}{\sigma - 1}$

Since $R = \frac{D}{2}$ and $r = \frac{d}{2}$, the ratio of $D/d$ is equal to the ratio of $R/r$, thus:

$\frac{D}{d} = \left(\frac{\sigma}{\sigma - 1}\right)^{1/3}$

This matches Option D:

$\left(\frac{\sigma}{\sigma-1}\right)^{1 / 3}$

$\begin{aligned} & v_\omega(1)+v_k(0.8)=\left(v_\omega+v_k\right)(0.9) \\ & \Rightarrow \frac{v_\omega}{v_k}=1 \end{aligned}$

To determine the percentage error in Young's modulus, we need to first understand the propagation of errors in the given formula.

Given the equation:

$ \mathrm{Y}=49000 \frac{\mathrm{M}}{\mathrm{l}} \frac{\mathrm{dyne}}{\mathrm{cm}^2} $

where:

• $M$ is the mass (with its value given as $500 \mathrm{~g}$)
• $l$ is the extension (with its value given as $2 \mathrm{~cm}$)

The errors in the measurements are determined by the smallest scale divisions on the graph paper, which are:

• $5 \mathrm{~g}$ for the load axis
• $0.02 \mathrm{~cm}$ for the extension axis

To find the percentage error in Young's modulus ($Y$), we need to compute the relative errors in the measurements $M$ and $l$, and then propagate these errors through the given formula.

The relative error in $M$ is:

$ \frac{\Delta M}{M} = \frac{5 \mathrm{~g}}{500 \mathrm{~g}} = 0.01 $

The relative error in $l$ is:

$ \frac{\Delta l}{l} = \frac{0.02 \mathrm{~cm}}{2 \mathrm{~cm}} = 0.01 $

Since $Y$ is proportional to $M$ and inversely proportional to $l$, the overall percentage error in $Y$ is the sum of the percentage errors in $M$ and $l$:

$ \frac{\Delta Y}{Y} = \frac{\Delta M}{M} + \frac{\Delta l}{l} = 0.01 + 0.01 = 0.02 $

To express this as a percentage, we multiply by 100:

$ \text{Percentage error in } Y = 0.02 \times 100 = 2\% $

Thus, the percentage error in Young's modulus $Y$ is:

Option A: 2%

Bernoulli's equation relates the pressure, velocity, and height in a flowing fluid and is derived from the principle of conservation of energy. The correct form of Bernoulli's equation is:

$P + \rho g h + \frac{1}{2} \rho v^2 = \text{constant}$

where:

• P is the pressure within the fluid.
• $\rho$ is the density of the fluid.
• g is the acceleration due to gravity.
• h is the height above a reference point.
• v is the velocity of the fluid.

Looking at the options given:

1. Option A: $P + \frac{1}{2} \rho g h + \frac{1}{2} \rho v^2 = \text{constant}$
2. Option B: $P + m g h + \frac{1}{2} m v^2 = \text{constant}$
3. Option C: $P + \rho g h + \rho v^2 = \text{constant}$
4. Option D: $P + \rho g h + \frac{1}{2} \rho v^2 = \text{constant}$

Option D correctly represents Bernoulli's equation in its proper form. Therefore, the correct answer is:

Option D

The difference in pressure inside a soap bubble as compared to the outside is due to the surface tension created by the soap film on the bubble. This difference in pressure can be calculated using the formula that relates the surface tension of the soap bubble to the radius of the bubble. The correct formula for the pressure difference ($\Delta P$) across a soap bubble is given by:

$\Delta P = \frac{4 S}{R}$

Here, $S$ is the surface tension of the bubble and $R$ is the radius of the bubble. The factor of 4 comes from the fact that a soap bubble has two surfaces (an inner and an outer surface), and for each surface, the Laplace pressure (which contributes to the pressure difference due to surface tension) is given by $\frac{2S}{R}$. Thus, for two surfaces, you double this amount, resulting in the $\frac{4 S}{R}$ term.

Therefore, the correct answer is:

Option B

$\frac{4 \mathrm{~S}}{\mathrm{R}}$

$\begin{aligned} & F_V=\left(m g-F_B\right) \frac{m g}{m g}=\left(\frac{\rho V_g-\rho_0 V_g}{\rho V_g}\right) m g \\ & F_v=m g\left(1-\frac{\rho_0}{\rho}\right) \end{aligned}$

To match List I with List II, we must understand the definitions or concepts described in List I and associate them with the correct terms in List II.

• A. A force that restores an elastic body of unit area to its original state: This definition describes Stress, as stress is the force applied over a unit area of a material that results in a deformation. Stress is involved in restoring the body to its original state.
• B. Two equal and opposite forces parallel to opposite faces: This is a description of Shear Modulus (also known as Modulus of Rigidity), which measures the material's response to shear stress (like forces applied parallel to one of its faces).
• C. Forces perpendicular everywhere to the surface per unit area same everywhere: This definition resembles the concept of Bulk Modulus, which is concerned with uniform pressure applied in all directions at points on the surface of a body and relates to volume change.
• D. Two equal and opposite forces perpendicular to opposite faces: This describes Young's Modulus (also known as Modulus of Elasticity), which involves tensile stress (force applied perpendicularly) leading to changes in length of a material.

Therefore, matching the lists results in the following:

• (A)-(III), because stress is the general concept applicable to the force per unit area described in restoring the body to its original state.
• (B)-(IV), because shear modulus is specifically related to parallel forces applied on opposite faces.
• (C)-(I), because bulk modulus deals with situations where force is applied uniformly over the surface of an object, affecting its volume.
• (D)-(II), because Young's modulus applies to forces perpendicular to the faces, affecting the length of a material.

Hence, the correct option is Option B (A)-(III), (B)-(IV), (C)-(I), (D)-(II).

Both statements given above have implications relating to the phenomena of capillarity, which involves the interaction between a liquid and a solid (in this case, a capillary tube). Let's break down each statement for clarity.

Statement I: When a capillary tube is dipped into a liquid and the liquid neither rises nor falls in the capillary, it suggests a scenario where the adhesive forces (between the liquid and the solid) and the cohesive forces (among the liquid molecules) are in perfect balance. The contact angle, which is the angle formed by the tangent to the liquid surface at the point of contact with the wall of the tube, plays a crucial role here. A contact angle of $0^{\circ}$ implies complete wetting, meaning the liquid spreads out to maximize contact with the solid. However, the statement that the liquid neither rises nor falls specifically with a contact angle of $0^{\circ}$ seems inaccurate. In reality, when the contact angle is $0^{\circ}$, it denotes perfect wetting, and the liquid tends to rise in the capillary tube. Thus, the precision of Statement I could be disputed based on the typical behavior of liquids in the context of capillarity and contact angles.

Statement II: The contact angle really is a property of both the solid and the liquid. It depends on the nature of the solid surface (whether it's hydrophilic or hydrophobic) and the type of liquid. This is because the contact angle reflects the degree of interaction between the liquid and solid surfaces, which is influenced by characteristics such as surface tension of the liquid and the surface energies of both the liquid and solid. Thus, Statement II accurately describes the nature of the contact angle as being dependent on both the material of the solid and the liquid.

Considering the analysis above:

• Statement I suggests a specific scenario that does not align accurately with the principles of capillarity and wettability, particularly at a contact angle of $0^{\circ}$, which typically would result in the liquid rising in the capillary, not staying neutral.
• Statement II accurately reflects the factors influencing the contact angle between a liquid and a solid surface.

Therefore, the correct option based on the given statements and their analysis would be:

Option B: Statement I is false but Statement II is true.

Option D, "Statement I is true but Statement II is false," is the correct choice. Here's an explanation for both statements:

Statement I: True

The contact angle between a solid and a liquid is indeed a measure of the wettability of the solid surface by the liquid. The contact angle is determined by the nature of both the solid and the liquid. It is a function of the interfacial tensions between solid-liquid ($ \gamma_{\text{SL}} $), solid-vapor ($ \gamma_{\text{SV}} $), and liquid-vapor ($ \gamma_{\text{LV}} $). This relationship can be understood through Young's equation:

$ \cos \theta = \frac{\gamma_{\text{SV}} - \gamma_{\text{SL}}}{\gamma_{\text{LV}}} $

Where $ \theta $ is the contact angle. Therefore, since the interfacial tensions vary with the materials in contact, the contact angle is indeed a property of the materials of both the solid and the liquid.

Statement II: False

The rise of a liquid in a capillary tube is strongly dependent on the inner radius of the tube. This relationship is described by the Jurin's Law, which states the height ($ h $) to which a liquid will rise (or fall) in a capillary tube is inversely proportional to the radius ($ r $) of the tube, among other factors. The law is given by:

$ h = \frac{2\gamma \cos \theta}{\rho g r} $

Where:

• $ \gamma $ is the liquid-air surface tension,
• $ \theta $ is the contact angle,
• $ \rho $ is the density of the liquid,
• $ g $ is the acceleration due to gravity,
• $ r $ is the radius of the capillary tube.

As seen from the equation, $ h $ is inversely proportional to $ r $. Therefore, the rise of the liquid indeed depends on the inner radius of the tube, making Statement II false.

If speed $=0$

Then $P_1+\rho g h_1=P_2+\rho g h_2$

In given ventury tube,

$\begin{aligned} & \qquad P_1+\rho g h+\frac{1}{2} \rho v_1^2=P_2+\frac{1}{2} \rho v_2^2 \\ & \Rightarrow \frac{1}{2} \rho\left(v_1^2-v_2^2\right)=\left(P_2-P_1\right)-\rho g h \\ & \Rightarrow v_1^2-v_2^2=\frac{2\left(P_2-P_1\right)}{\rho}-2 g h \end{aligned}$

To answer this question, we need to understand the relationship between the surface area of the droplets and the surface energy involved.

Surface energy is directly proportional to the surface area of the liquid. The surface energy, $ E $, for a droplet is given by:

$ E = \gamma \times A $

Where:

• $ E $ is the surface energy,
• $ \gamma $ is the surface tension of the liquid, and
• $ A $ is the surface area of the droplet.

When multiple droplets coalesce, they form a larger droplet with a certain volume. Since the volume is conserved, the volume of the large droplet will be equal to the sum of the volumes of the small droplets.

Let's denote:

• $ r $ as the radius of a small droplet,
• $ R $ as the radius of the large droplet,
• $ V_{\text{small}} $ as the volume of a small droplet, and
• $ V_{\text{large}} $ as the volume of the large droplet.

The volume of one small droplet is:

$ V_{\text{small}} = \frac{4}{3}\pi r^3 $

The total volume of 1000 small droplets is:

$ 1000 \times V_{\text{small}} = 1000 \times \frac{4}{3}\pi r^3 $

Since the volume is conserved, the volume of the large droplet formed by the coalescence of 1000 small droplets is:

$ V_{\text{large}} = 1000 \times \frac{4}{3}\pi r^3 $

Now, if $ R $ is the radius of the large droplet, then:

$ V_{\text{large}} = \frac{4}{3}\pi R^3 $

Equating the volumes, we have:

$ \frac{4}{3}\pi R^3 = 1000 \times \frac{4}{3}\pi r^3 $

$ R^3 = 1000 \times r^3 $

$ R = 10r $

Now, let's look at the surface area. The surface area for a small droplet is $ A_{\text{small}} = 4\pi r^2 $ and for a large droplet is $ A_{\text{large}} = 4\pi R^2 $. Substitute $ R = 10r $:

$ A_{\text{large}} = 4\pi (10r)^2 $

$ A_{\text{large}} = 4\pi \times 100r^2 $

$ A_{\text{large}} = 100 \times 4\pi r^2 $

$ A_{\text{large}} = 100 \times A_{\text{small}} $

So, the large droplet has 100 times the surface area of one small droplet.

The surface energy of 1000 small droplets would be $ 1000 \times E_{\text{small}} $ because each small droplet has an energy $ E_{\text{small}} = \gamma \times A_{\text{small}} $.

The surface energy of the big droplet is $ E_{\text{large}} = \gamma \times A_{\text{large}} $. But we have just shown that $ A_{\text{large}} = 100 \times A_{\text{small}} $, so:

$ E_{\text{large}} = \gamma \times 100 \times A_{\text{small}} $

This means the surface energy of the big droplet is 100 times the surface energy of one small droplet. Since there were 1000 small droplets originally, the surface energy of the big droplet is:

$ \frac{E_{\text{large}}}{1000 \times E_{\text{small}}} = \frac{\gamma \times 100 \times A_{\text{small}}}{1000 \times \gamma \times A_{\text{small}}} = \frac{1}{10} $

Therefore, the correct answer is:

Option B: The surface energy will become $\frac{1}{10}$ th of the original.

The Young's modulus of elasticity, denoted as $E$, is a measure of the stiffness of a material. It defines the relationship between stress (force per unit area) and strain (deformation) in a material in the linear (elastic) portion of the stress-strain curve. As temperature changes, the interatomic distances and bonding energies within a material also change, affecting its mechanical properties, including its elasticity.

For most materials, as the temperature increases, the atoms within the material gain kinetic energy and vibrate more vigorously. This increased vibration results in a reduction of the forces between atoms, making it easier for the material to deform under a given load. Hence, generally, the Young's modulus $E$ decreases with increasing temperature because the material becomes softer and less stiff. Thus, the correct option here would be:

Option C: decreases

However, it should be noted that the exact relationship between temperature and Young's modulus can vary depending on the material type and its microstructure. While the general trend for metals and polymers is a decrease in Young's modulus with rising temperature, the rate of decrease and the temperature range over which this occurs can differ significantly between materials. Some advanced materials and composites may exhibit more complex behavior due to their unique properties.

Since density is negligible hence Buoyancy force will be negligible

At terminal velocity.

$\mathrm{Mg} =6 \pi \eta \mathrm{rv}$

$\mathrm{V} \propto \frac{1}{\mathrm{r}} \quad$ (as mass is constant)

Now, $\frac{\mathrm{v}}{\mathrm{v}^{\prime}}=\frac{\mathrm{r}^{\prime}}{\mathrm{r}}$

$r^{\prime}=2 \mathrm{r}$

So, $v^{\prime}=\frac{v}{2}$

$\begin{aligned} & \mathrm{mg}-\mathrm{F}_{\mathrm{B}}-\mathrm{F}_{\mathrm{v}}=\mathrm{ma} \\ & \left(\rho \frac{4}{3} \pi \mathrm{r}^3\right) g-\left(\rho_{\mathrm{L}} \frac{4}{3} \pi \mathrm{r}^3\right) g-6 \pi \eta r v=m \frac{d v}{d t} \end{aligned}$

Let $\frac{4}{3 m} \pi R^3 g\left(\rho-\rho_L\right)=K_1$ and $\frac{6 \pi \eta r}{m}=K_2$

$\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{K}_1-\mathrm{K}_2 \mathrm{~V}$

$\begin{aligned} & \int_\limits0^v \frac{d v}{K_1-K_2 v}=\int_\limits0^t d t \\ & -\frac{1}{K_2} \ln \left[K_1-K_2 v\right]_0^v=t \\ & \ell n\left(\frac{K_1-K_2 v}{K_1}\right)=-K_2 t \\ & K_1-K_2 v=K_1 e^{-K_2 t} \\ & v=\frac{K_1}{K_2}\left[1-e^{-K_2 t}\right] \end{aligned}$

Young's modulus depends on the material not length and cross sectional area. So young's modulus remains same.

Volume constant

$\begin{aligned} & \frac{4}{3} \pi R^3=27 \times \frac{4}{3} \times \pi r^3 \\ & R^3=27 r^3 \\ & R=3 r \\ & r=\frac{R}{3} \\ & r^2=\frac{R^2}{9} \end{aligned}$

$\begin{aligned} & \text { Work done }=T . \Delta A \\ & =27 T\left(4 \pi r^2\right)-T 4 \pi R^2 \\ & =27 T 4 \pi \frac{R^2}{9}-4 \pi R^2 T \\ & =8 \pi R^2 T \end{aligned}$

$\begin{aligned} & Y=\frac{\text { stress }}{\text { strain }} \\ & Y=\frac{\frac{\mathrm{F}}{\frac{\pi \mathrm{r}^2}{\ell}}}{\frac{\ell}{\mathrm{L}}} \end{aligned}$

$\mathrm{F}=\mathrm{Y} \pi \mathrm{r}^2 \times \frac{\ell}{\mathrm{L}} \quad \text{.... (i)}$

$\begin{aligned} & \mathrm{Y}=\frac{\frac{\mathrm{F} / 2}{\pi \mathrm{r}^2 / 4}}{\frac{\Delta \ell}{\mathrm{L}}} \\ & \mathrm{F}=\mathrm{Y} \frac{\Delta \ell}{\mathrm{L}} \times 2 \times \frac{\pi \mathrm{r}^2}{4} \end{aligned}$

From (i)

$\begin{aligned} & \mathrm{Y} \pi \mathrm{r}^2 \frac{\ell}{\mathrm{L}}=\mathrm{Y} \frac{\Delta \ell}{\mathrm{L}} \frac{\pi \mathrm{r}^2}{2} \\ & \Delta \ell=2 \ell \end{aligned}$

Surface tension will be less as temperature increases

$\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}}$

Height of capillary rise will be smaller in hot water and larger in cold water.